Question:

A crystalline solid $X$ reacts with $dil. HCl$ to liberate a gas $Y.Y$ decolourises acidified $KMnO_4$. When a gas $'Z'$ is slowly passed into an aqueous solution of $Y$, colloidal sulphur is obtained. $X$ and $Z$ could be, respectively

Updated On: May 15, 2024
  • $Na_2SO_4,H_2S$
  • $Na_2SO_4,SO_2$
  • $Na_2S,SO_3$
  • $Na_2SO_3,H_2S$
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The Correct Option is D

Solution and Explanation

$Na _{2} SO _{3}(X)$ reacts with dil. $HCl$ to liberate $SO _{2}$ gas $(Y)$.
This $SO _{2}$ gas decolourises acidified $KMnO _{4} .$ On passing $H _{2} S$ gas $(Z)$ into aqueous solution of $SO _{2}$ (i.e., $H _{2} SO _{3}$ ), colloidal sulphur is obtained.
Complete reactions are as follows

$\underset{\text{Purple}}{2KMnO_4} + 2H_2O + \underset{(Y)}{5SO_2} \longrightarrow K_2SO_4 + \underset{\text{Colourless}}{2MnSO_4} + 2H_2SO_4$
$H _{2} O +\underset{(Y)}{ SO _2} \longrightarrow \underset{\text{Aqueous solution of Y}}{H_2SO_3}$
$H _{2} SO _{3}+\underset{(Z)}{2 H _{2} S} \longrightarrow \underset{\text{Colloidal sulphur}}{2 S +3 H _{2} O}$
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